1. **State the problem:** We have two parabolas given by the equations: $$y = x^2 - c^2$$ and $$y = c^2 - x^2$$ These parabolas bound a region whose area is 1944. We need to find the value of $c$. 2. **Understand the region bounded:** The region bounded by these curves is where the two parabolas intersect. Since one is $y = x^2 - c^2$ (opening upwards shifted down by $c^2$) and the other is $y = c^2 - x^2$ (opening downwards shifted up by $c^2$), the region is symmetric about the y-axis. 3. **Find the points of intersection:** Set the two equations equal: $$x^2 - c^2 = c^2 - x^2$$ Simplify: $$x^2 - c^2 = c^2 - x^2$$ $$x^2 + x^2 = c^2 + c^2$$ $$2x^2 = 2c^2$$ $$\cancel{2}x^2 = \cancel{2}c^2$$ $$x^2 = c^2$$ So, $$x = \pm c$$ 4. **Set up the integral for the area:** The area between the curves from $x = -c$ to $x = c$ is: $$A = \int_{-c}^c \big[(c^2 - x^2) - (x^2 - c^2)\big] \, dx$$ Simplify the integrand: $$(c^2 - x^2) - (x^2 - c^2) = c^2 - x^2 - x^2 + c^2 = 2c^2 - 2x^2$$ 5. **Calculate the integral:** $$A = \int_{-c}^c (2c^2 - 2x^2) \, dx = 2 \int_{-c}^c (c^2 - x^2) \, dx$$ Because the function is even, we can write: $$A = 2 \times 2 \int_0^c (c^2 - x^2) \, dx = 4 \int_0^c (c^2 - x^2) \, dx$$ 6. **Evaluate the integral:** $$\int_0^c (c^2 - x^2) \, dx = \left[c^2 x - \frac{x^3}{3}\right]_0^c = c^3 - \frac{c^3}{3} = \frac{2c^3}{3}$$ 7. **Calculate total area:** $$A = 4 \times \frac{2c^3}{3} = \frac{8c^3}{3}$$ 8. **Use the given area to solve for $c$:** $$\frac{8c^3}{3} = 1944$$ Multiply both sides by 3: $$8c^3 = 1944 \times 3 = 5832$$ Divide both sides by 8: $$c^3 = \frac{5832}{8} = 729$$ 9. **Find $c$:** $$c = \sqrt[3]{729} = 9$$ **Final answer:** $$\boxed{9}$$